What is the integral of (e^x + 1)/(e^x - 1)?

The integral of

\frac{e^x + 1}{e^x - 1}

2\ln\lvert e^x - 1 \rvert - x + C

.

Solution. We want to determine the integral of

\frac{e^x + 1}{e^x - 1}

, i.e.:

\begin{align*} \int \frac{e^x + 1}{e^x - 1} dx = \int \frac{e^x}{e^x - 1} dx + \int \frac{1}{e^x - 1} dx. \end{align*}

First we will determine

\int \frac{e^x}{e^x - 1} dx

. We will use the substitution method. So take

u = e^x - 1

. Then we saw here that

du = e^x dx

. Therefore, we get the following:

\begin{align*} \int \frac{e^x}{e^x - 1} dx &= \int \frac{du}{u} \\ &= \ln \lvert u \rvert + C_1 \\ &= \ln \lvert e^x - 1 \rvert + C_1, \end{align*}

where

C_1

is some constant. Now what is left is to determine

\int \frac{1}{e^x - 1} dx

. This integral is difficult to determine, but we make it easy by using the following handy adjustment:

\begin{align*} \int \frac{1}{e^x - 1} dx = \int \frac{1}{e^x(1 - e^{-x})} dx = \int \frac{e^{-x}}{1 - e^{-x}} dx. \end{align*}

Now we will apply the substitution method. Let

v = 1 - e^{-x}

. Then we saw here that

dv = e^{-x} dx

. So we get the following:

\begin{align*} \int \frac{e^{-x}}{1 - e^{-x}} dx &= \int \frac{dv}{v} \\ &= \ln \lvert v \rvert + C_2 \\ &= \ln \lvert 1 - e^{-x} \rvert + C_2 \\ &= \ln \lvert e^{-x}(e^x - 1) \rvert + C_2 \\ &= \ln \lvert e^{-x} \rvert + \lvert e^{-x} - 1 \rvert + C_2 \\ &= -x + \lvert e^{-x} - 1 \rvert + C_2, \end{align*}

where

C_2

is some constant. Now we will combine everything together:

\begin{align*} \int \frac{e^x + 1}{e^x - 1} dx &= \int \frac{e^x}{e^x - 1} dx + \int \frac{1}{e^x - 1} dx \\ &= \ln \lvert e^x - 1 \rvert + C_1 -x + \lvert e^{-x} - 1 \rvert + C_2 \\ &= 2\ln\lvert e^x - 1 \rvert - x + C, \end{align*}

where

C = C_1 + C_2

. So, the integral of

\frac{e^x + 1}{e^x - 1}

2\ln\lvert e^x - 1 \rvert - x + C

What is the integral of (e^x + 1)/(e^x – 1)?